Bergman Kernel and Pluripotential Theory

Bergman Kernel and Pluripotential Theory

Zbigniew B locki

Uniwersytet Jagiello´ nski, Krak´ ow, Poland http://gamma.im.uj.edu.pl/ eblocki

Geometric invariance and nonlinear PDE

Kioloa, February 9-14, 2014

Bergman Completeness

Ω bounded domain in C

H

(Ω) = O(Ω) ∩ L

(Ω) K

(·, ·) Bergman kernel

f (w ) = Z

Ω

f K

(·, w ) d λ, w ∈ Ω, f ∈ H

(Ω)

K

(w ) = K

(w , w )

= sup{|f (w )|

: f ∈ H

(Ω), ||f || ≤ 1}

Ω is called Bergman complete if it is complete w.r.t. the Bergman

metric B

= i ∂ ¯ ∂ log K

Kobayashi Criterion (1959) If

w →∂Ω

lim

|f (w )|

K

(w ) = 0, f ∈ H

(Ω), then Ω is Bergman complete.

The opposite is not true even for n = 1 (Zwonek, 2001).

Kobayashi Criterion easily follows using the embedding ι : Ω 3 w 7−→ [K

(·, w )] ∈ P(H

(Ω)) and the fact that ι

ω

= B

.

Since ι is distance decreasing,

dist

(z, w ) ≥ arccos |K

(z, w )|

pK

(z)K

(w ) .

Some Pluripotential Theory

Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u ∈ PSH

(Ω) s.th. u = 0 on ∂Ω).

Demailly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex.

Pluricomplex Green function For a pole w ∈ Ω we set

G

(·, w ) = G

= sup{v ∈ PSH

(Ω) : v ≤ log | · −w | + C } Lempert (1981) Ω convex ⇒ G

symmetric

Demailly (1985) Ω hyperconvex ⇒ e

∈ C ( ¯ Ω × Ω) Open Problem e

∈ C ( ¯ Ω × ¯ Ω \ ∆

)

Equivalently: G (·, w

) → 0 loc. uniformly as w

→ ∂Ω?

True if ∂Ω ∈ C

(Herbort, 2000)

Demailly (1985) If Ω is hyperconvex then G

= G

(·, w ) is the unique solution to



 

 



 

 



u ∈ PSH(Ω) ∩ C ( ¯ Ω \ {w }) (dd

u)

= (2π)

δ

u = 0 on ∂Ω u ≤ log | · −w | + C

B. (1995) If Ω is hyperconvex then ∃! u = u

s.th.



 

 

u ∈ PSH(Ω) ∩ C ( ¯ Ω) (dd

u)

= 1 d λ u = 0 on ∂Ω.

Open Problem u ∈ C

(Ω)

Pogorelov (1971) True for the analogous solution of the real

Monge-Amp` ere equation (for any bounded convex domain in R

without any regularity assumptions).

B.-Y. Chen, Pflug - B. (1998) / Herbort (1999) Hyperconvex domains are Bergman complete Herbort If Ω is pseudoconvex then

|f (w )|

K

(w ) ≤ c

Z

{Gw<−1}

|f |

d λ, w ∈ Ω, f ∈ H

(Ω).

Corollary lim

w →∂Ω

λ({G

< −1}) = 0 ⇒ Ω is Bergman complete Proposition If Ω is hyperconvex then

w →∂Ω

lim ||G

||

= 0.

Sketch of proof ||G

||

= R

Ω

|G

|

(dd

u

)

≤ n!||u

||

Z

Ω

|u

|(dd

G

)

≤ C (n, λ(Ω)) |u

(w )|

Lower Bound for the Bergman Distance

Diederich-Ohsawa (1994), B. (2005) If Ω is pseudoconvex with C

boundary then

dist

(·, w ) ≥ log δ

C log log δ

, where δ

(z) = dist

(z, ∂Ω).

Pluripotential theory is the main tool in proving this estimate, in particular we have the following:

B. (2005) If Ω is pseudoconvex and z, w ∈ Ω are such that {G

< −1} ∩ {G

< −1} = ∅

then

dist

(z, w ) ≥ c

> 0.

Open Problem dist

(·, w ) ≥

log δ

From Herbort’s estimate

|f (w )|

K

(w ) ≤ c

Z

{Gw<−1}

|f |

d λ, w ∈ Ω, f ∈ H

(Ω),

for f ≡ 1 we get

K

(w ) ≥ 1

c

λ({G

< −1}) .

To find the optimal constant c

here turns out to have very interesting consequences!

Herbort (1999) c

= 1 + 4e

, where Ω ⊂ B(z

, R) (Main tool: H¨ ormander’s estimate for ¯ ∂) B. (2005) c

= (1 + 4/Ei (n))

, where Ei (t) =

Z

ds se

(Main tool: Donnelly-Fefferman’s estimate for ¯ ∂)

Suita Conjecture

D bounded domain in C c

(z) := exp lim

ζ→z

(G

(ζ, z) − log |ζ − z|)

(logarithmic capacity of C \ D w.r.t. z) c

|dz| is an invariant metric (Suita metric)

Curv

= − (log c

)

c

Suita Conjecture (1972): Curv

≤ −1

• “=” if D is simply connected

• “<” if D is an annulus (Suita)

• Enough to prove for D with smooth boundary

• “=” on ∂D if D has smooth boundary

-5 -4 -3 -2 -1

-7 -6 -5 -4 -3 -2 -1

Curv

for D = {e

< |z| < 1} as a function of log |z|

-5 -4 -3 -2 -1

-6 -5 -4 -3 -2 -1

Curv

for D = {e

< |z| < 1} as a function of log |z|

∂

∂z∂¯ z (log c

) = πK

(Suita) Therefore the Suita conjecture is equivalent to

c

≤ πK

.

Ohsawa (1995) observed that it is really an extension problem: for z ∈ D find holomorphic f in D such that f (z) = 1 and

Z

D

|f |

d λ ≤ π (c

(z))

.

Using the methods of the Ohsawa-Takegoshi extension theorem he showed the estimate

c

≤ C πK

with C = 750.

C = 2 (B., 2007)

C = 1.95388 . . . (Guan-Zhou-Zhu, 2011)

Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013)

0 ∈ D ⊂ C, Ω ⊂ C

× D, Ω pseudoconvex, ϕ ∈ PSH(Ω)

f holomorphic in Ω

:= Ω ∩ {z

= 0}

Then there exists a holomorphic extension F of f to Ω such that Z

Ω

|F |

e

d λ ≤ π c

(0)

Z

Ω⁰

|f |

e

d λ

.

For n = 1 and ϕ ≡ 0 we get the Suita conjecture.

Main tool: H¨ ormander’s estimate for ¯ ∂

B.-Y. Chen (2011) proved that the Ohsawa-Takegoshi theorem

(without optimal constant) follows form H¨ ormander’s estimate.

Tensor Power Trick

We have

K

(w ) ≥ 1

c

λ({G

< −1}) where c

= (1 + 4/Ei (n))

.

Take m  0 and set e Ω := Ω

⊂ C

, w := (w , . . . , w ). Then e K

( w ) = (K e

(w ))

, λ

({G

< −1}) = (λ

({G

< −1})

. Therefore

K

(w ) ≥ 1

c

λ({G

< −1}) but

m→∞

lim c

= e

.

Repeating this argument for any sublevel set we get

Theorem 1 Assume Ω is pseudoconvex in C

. Then for a ≥ 0 and w ∈ Ω

K

(w ) ≥ 1

e

λ({G

< −a}) .

Lempert recently noticed that this estimate can also be proved using Berndtsson’s result on positivity of direct image bundles.

What happens when a → ∞?

For n = 1 we get K

≥ c

/π (another proof of Suita conjecture).

For n ≥ 1 and Ω convex using Lempert’s theory one can obtain:

Theorem 2 If Ω is a convex domain in C

then for w ∈ Ω K

(w ) ≥ 1

λ(I

(w )) ,

I

(w ) = {ϕ

(0) : ϕ ∈ O(∆, Ω), ϕ(0) = w } (Kobayashi indicatrix).

Mahler Conjecture

K - convex symmetric body in R

K

:= {y ∈ R

: x · y ≤ 1 for every x ∈ K } Mahler volume := λ(K )λ(K

)

Mahler volume is an invariant of the Banach space defined by K : it is independent of linear transformations and of the choice of inner product.

Santal´ o Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes Hansen-Lima bodies: starting from an interval they are produced by taking products of lower dimensional HL bodies and their duals.

n = 2: square

n = 3: cube & octahedron

n = 4: . . .

Bourgain-Milman (1987) There exists c > 0 such that λ(K )λ(K

) ≥ c

4

n! . Mahler Conjecture: c = 1

G. Kuperberg (2006) c = π/4 Nazarov (2012)

I

equivalent SCV formulation of the Mahler Conjecture via the Fourier transform and the Paley-Wiener theorem

I

proof of the Bourgain-Milman Inequality (c = (π/4)

) using

H¨ ormander’s estimate for ¯ ∂

K - convex symmetric body in R

Nazarov: consider T

:= intK + i R

⊂ C

. Then

 π 4



1

(λ

(K ))

≤ K

(0) ≤ n!

π

λ

(K

) λ

(K ) . Therefore

λ

(K )λ

(K

) ≥  π 4



4

n! . To show the lower bound we can use Theorem 2:

K

(0) ≥ 1 λ

(I

(0)) .

Proposition I

(0) ⊂ 4

π (K + iK ) In particular, λ

(I

(0)) ≤  4

π



(λ

(K ))

Conjecture λ

(I

(0)) ≤  4 π



(λ

(K ))